5x+7 1. If then (x+1)(x+2) x+1 x+2 D. a = 12, b = 17 A. a = 2, b = 3 B. a = 3, b= 2 C. a 2, b= -3 4x+1 2. Expressed in partial fractions 2x2 +x-3 2x 1 - D. 1 4 A. - 2 1 7 В. 5(2x+3) 2x+3 x-1 2x+3 X-1 5(x-1) 2x + 3 x-1 the values of the constants P and Q are given by; X-3 5x P Q 3. Given that (x+2)(x-3) x+2 A. P = 3, Q = -2 B. P = 2,Q = -3 C. P = -2,Q = -3 D. P = 2, Q = 3 5x+4 P 4. If then (2x+1)(3x+2) 2x+1 3x+2' A. P = 3, Q = 2 B. P= -2, Q= 3 C. P= 3, Q=-2 D.P-1, Q = 2 5. (x+1)(x-2) +C. 1+ 3(x+1) 1 4 1 1 4 A. 3(x-2) 1 В. 1 3(x+1) D.1- X+1 x-2 3(x-2) 3(x+1) 3(x-2) 5x + 3 6. (x-1)(x2+1) 4 - B.- 1 2 1+x x+1 4 D. + x2+1 x-1 A. X-1 x2+1 X-1 1+x2 x-1 x²+1 x-1
5x+7 1. If then (x+1)(x+2) x+1 x+2 D. a = 12, b = 17 A. a = 2, b = 3 B. a = 3, b= 2 C. a 2, b= -3 4x+1 2. Expressed in partial fractions 2x2 +x-3 2x 1 - D. 1 4 A. - 2 1 7 В. 5(2x+3) 2x+3 x-1 2x+3 X-1 5(x-1) 2x + 3 x-1 the values of the constants P and Q are given by; X-3 5x P Q 3. Given that (x+2)(x-3) x+2 A. P = 3, Q = -2 B. P = 2,Q = -3 C. P = -2,Q = -3 D. P = 2, Q = 3 5x+4 P 4. If then (2x+1)(3x+2) 2x+1 3x+2' A. P = 3, Q = 2 B. P= -2, Q= 3 C. P= 3, Q=-2 D.P-1, Q = 2 5. (x+1)(x-2) +C. 1+ 3(x+1) 1 4 1 1 4 A. 3(x-2) 1 В. 1 3(x+1) D.1- X+1 x-2 3(x-2) 3(x+1) 3(x-2) 5x + 3 6. (x-1)(x2+1) 4 - B.- 1 2 1+x x+1 4 D. + x2+1 x-1 A. X-1 x2+1 X-1 1+x2 x-1 x²+1 x-1
Chapter7: Systems Of Equations And Inequalities
Section7.4: Partial Fractions
Problem 5SE: Once you have a system of equations generated by the partial fraction decomposition, can you explain...
Related questions
Concept explainers
Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
Solve Q7, & 8 showing clearly all steps involved
Transcribed Image Text:5x+7
b
then
x+2
a
1. If
(x+1)(x+2)
X+1
%3D
A. a = 2, b= 3 B. a = 3, b= 2 C. a = 2, b= -3 D. a = 12, b = 17
%3D
%3D
4x+1
2.
Expressed in partial fractions
2x2 +x-3
1
2x
D.
1
4
2
1
A.
2x+3
В.
5(2x+3)
2x+3
X-1
X-1
5(x-1)
2x + 3
X-1
5x
P
3. Given that
the values of the constants P and Q are given by;
(x+2)(x-3)
x+2
X-3
A. P = 3, Q = -2 B. P = 2, Q = -3 C. P = -2, Q = -3
D. P = 2, Q = 3
|
5x+4
4. If
then
(2x+1)(3x+2)
2x+1
3x+2'
A. P = 3, Q = 2 B. P= -2, Q= 3 C. P= 3, Q=-2 D.P=1, Q= 2
5.
(x+1)(x-2)
4
1
1
4.
4
1
4
A.
3(x-2)
B. 1-
+ C. 1+
D. 1
-
3(x+1)
X-2
3(x+1)
3 (х-2)
3(x+1)
3(x-2)
5x? + 3
6.
(x-1)(x²+1)
A. +B.-d+
1+x
4
X+1
4
D.
4
X-1
x2+1
1+x2
X-1
x²+1
x² +1
X-1
X-1
X-1
then
m
7. If
- =
x2 -4
X-2
X+2'
1
1
1
B. m=
C. m=2, n=2 D. m-
1
A. m
2
2
4
x2 +3
8.
x(x2 +2)
3.
A.
2x
3
В.
2x
3.
D.
2(x2 +2)
2(x²+2)
2 (x2 +2)
2(x2 +2)
2x
2x
x2 - 1
9.
(2x-1)2
X-1
В.
x2
Ç D.
2
1
3
1
x-2
A.-
x2
2x+1
2x+1
2x+1
2x+1
x+1
are
10. The partial fractions of
2:-2
is. 1 +
e.1+
3
3
A. 1+
C. 1+
D.
X+1
X-2
X-2
X-2
x-2
20
3.
3.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 1 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.Recommended textbooks for you
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
College Algebra
Algebra
ISBN:
9781305115545
Author:
James Stewart, Lothar Redlin, Saleem Watson
Publisher:
Cengage Learning
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
College Algebra
Algebra
ISBN:
9781305115545
Author:
James Stewart, Lothar Redlin, Saleem Watson
Publisher:
Cengage Learning